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(B) The sum of all probabilities in a probability distribution is $1$. Therefore, $\sum_{r=0}^{\infty} P(X=r) = 1$. Given $P(X=r) = k(1+r) 3^{-r}$, we

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$\frac{8}{9}$

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(B) The sum of all probabilities in a probability distribution is $1$. Therefore, $\sum_{r=0}^{\infty} P(X=r) = 1$. Given $P(X=r) = k(1+r) 3^{-r}$, we have $k \sum_{r=0}^{\infty} (1+r) \left(\frac{1}{3}ight)^r = 1$. Let $S = \sum_{r=0}^{\infty} (1+r) x^r$ where $x = \frac{1}{3}$. This is an arithmetico-geometric series: $S = 1 + 2x + 3x^2 + 4x^3 + \ldots$. Multiplying by $x$: $xS = x + 2x^2 + 3x^3 + \ldots$. Subtracting the two: $S(1-x) = 1 + x + x^2 + x^3 + \ldots = \frac{1}{1-x}$. Thus, $S = \frac{1}{(1-x)^2}$. For $x = \frac{1}{3}$, $S = \frac{1}{(1 - 1/3)^2} = \frac{1}{(2/3)^2} = \frac{9}{4}$. So, $k 	imes \frac{9}{4} = 1 \implies k = \frac{4}{9}$. Now, $P(X=0) + P(X=1) + P(X=2) = k \left[ (1+0)3^0 + (1+1)3^{-1} + (1+2)3^{-2} ight]$. $= \frac{4}{9} \left[ 1 + \frac{2}{3} + \frac{3}{9} ight] = \frac{4}{9} \left[ 1 + \frac{2}{3} + \frac{1}{3} ight] = \frac{4}{9} 	imes 2 = \frac{8}{9}$.

$A$ random variable $X$ has the range $\{0, 1, 2, \ldots\}$. If $P(X=r) = k(1+r) 3^{-r}$ for $r=0, 1, 2, \ldots$, where $k > 0$ is a real number, then $P(X=0) + P(X=1) + P(X=2) =$

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